The Resistor Cube Equivalent Resistance Conundrum

You have probably seen somewhere along the line in your electronics career the resistor cube problem.
The 12 edges of the cube each contain a 1 Ω resistor, and the challenge is to calculate what the equivalent
resistance is between two opposing corners. It is a daunting problem using straight circuit analysis,
since it requires writing and solving multiple mesh equations. There are lots of opportunities for making
mistakes.

One option if you had the time and facilities would be to build the model in a circuit simulator
and let it determine the result. Usually, though, the cube is thrust upon you in a compromising situation,
like in a job interview. If you are an electrical engineer and cannot figure it out on the spot, forget
that circuit design job. If you are an electronics technician, you will be forgiven for not solving
it, but you had better demonstrate an understanding of the method once it is presented.

As it turns out, there is a relatively simple analysis based on symmetry and a fundamental level
of understanding of currents and voltages.

The traditional method used involves recognizing
sets of equipotential points within the vertices of the cube, then shorting them together to enable
calculation of parallel resistances. Finally, those resistances are added in series to arrive at the
resulting equivalent resistance. The process is illustrated below.

After explaining the traditional method, I will present my
solution, which is a little more intuitive and direct method for arriving at the same answer. Solving
via the traditional method actually requires the same knowledge of how currents are divided at nodes.

This is the cube structure consisting of 12 resistors electrically connected between the 8 vertices.
Each resistor is 1 Ω, but any value can be used so long as they are all the same.

Here is where the intuition comes into play. Color coding is used to help keep track of the resistors
and associated nodes (below). Due to symmetry, the potential (voltage) at the three nodes labeled "α" are equal. Since no current flows between nodes with a potential difference
of 0 V, they can be shorted together without affecting the circuit's integrity. The same can be done
for the nodes labeled "β."

Once you short those nodes, you obtain the equivalent circuit shown below. As you can see, there
are two sets of three resistors in parallel, in series with one set of six resistors in parallel. So,
you have 1/3 Ω in series with 1/6 Ω in series with 1/3 Ω, which equals 5/6 Ω.

Now I will present my method of solving the resistor cube problem. The structure is repeated again here.

Kirchhoff's current law, which states that the sum of the currents entering and exiting a node is
zero, is essential in the analysis.

The first step is to recognize that at a node where equal resistances exist, current entering the
node will be distributed equally between the number of output branches - in this case three. For convenience
sake, I assigned an input current of 3 amperes at the corner labeled "A," so that 1 amp will flow through
each output branch. Note that 1 A flows through each branch.

On the far side of each of those branches is another node with two output branches. Again, due to
symmetry, the input current will divide evenly so that ½ A flow into each branch. Looking at the cube's
output node labeled "B," it is apparent that the same situation exists as with "A."

Take a moment to sum the currents into and out of each node to verify that they all add up as required.

Now that you know the current through each branch, and you know that each branch has a single 1 Ω
resistor in it, Ohms law allows you to calculate the voltage across each resistor.

The next step is to sum the voltage from input node "A" to output node "B." Any path you take travels
along three edges, and all total to 2½ volts.

Finally, apply Ohms law, which says that the resistance is equal to the voltage divided by the current.
As with the other analysis method, the resulting equivalent resistance is 5/6 Ω.

You can see that in reality, being able to make the assumptions in the traditional solution requires
an understanding of the current division principles in my method. So, IMHO it is simpler to add voltages
and then plug voltage and current into Ohm's law to arrive at the answer than to risk shorting nodes
incorrectly. It's so simple, a caveman could do it.

As a verification of the result, the resistor cube circuit was simulated using the free
LTSpice program,
by Linear Technology. Resistors are labeled in accord with the labels in the
traditional method of analysis. A 3 amp current source
is placed at input node N001. The resulting voltage is the predicted 2.5 V. Again,
Ohm's law for 3 amps and 2.5 volts yields a resistance of 5/6 Ω.

Thanks to RF Cafe visitor Les Carpenter for sending me
this solution that rearranges the resistors in delta and star configurations in order to "simplify"
the solution. My head is still hurting from looking at it.

Posted June 27, 2010

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